An unknown element "X" has electron configuration ["Ar"]3d^5 and a charge of 3+. What is the atomic number of "X"?

1 Answer
Oct 1, 2017

26

Explanation:

For starters, you know that your element is located in period 4 of the Periodic Table because all the transition metals that have 3d configurations are located in period 4 of the Periodic Table.

This means that you can write the noble-gas shorthand configuration for this unknown element, let's say "X", like this

"X"^(3+): ["Ar"] 3d^5

Now, the trick here is to realize that when a transition metal located in period 4 is losing electrons, the first two electrons that are lost are coming from the 4s orbital.

In other words, when the 4s orbital is occupied, it is actually higher in energy than the 3d orbitals. Keep in mind that when the 4s orbital is empty, it is actually lower in energy than the 3d orbitals!

So when "X" is losing the first two electrons, they are being taken from the 4s orbital. Consequently, you can say that when the element is losing the third electron, this electron will come from the 3d orbitals.

You can thus backtrack and add electrons to the 3+ cations to form the neutral atom.

"X"^(2+) = "X"^(3+) + "e"^(-)

This means that you have--remember, the third electron was removed from the 3d orbitals, so it must be added back to the 3d orbitals!

"X"^(2+): ["Ar"] 3d^6

Next, you have

"X" = "X"^(2+) + 2"e"^(-)

This means that you have--the first two electrons were removed from the 4s orbital, so they must be added back to the 4s orbital!

"X": ["Ar"]3d^6 4s^2

Therefore, you can say that a neutral atom of "X" has a total of

overbrace("18 e"^(-))^(color(blue)("the same as neutral atom of Ar")) + overbrace("6 e"^(-))^(color(blue)("in the 3d orbitals")) + overbrace("2 e"^(-))^(color(blue)("in the 4s orbitals")) = "26 e"^(-)

This implies that a neutral atom of "X" must have 26 protons located inside its nucleus.

You can conclude that your unknown element is iron, "Fe", which has an atomic number equal to 26.